Re: [algogeeks] seperate diff types of coins

2012-07-10 Thread Navin Kumar
Minimum no. weighings:

Divide 8 coins in group of 3, 3 and 2.

For minimum weighsing group1 's total weight is x units(say) --FIrst
weighing
Groups 2nd total weights is y units  Second weighing.
Lastly one more weighing among a unit and b unit coins.---3 rd weighing


So minimum 3 weighing is required.



On Tue, Jul 10, 2012 at 11:03 AM, payal gupta gpt.pa...@gmail.com wrote:

 You have 8 coins. 3 of them weigh x units, 3 y units, 1 a units and 1 b
 units. They are all mixed and look identical. What are the minimum no of
 weighings reqd to seperate the for types of coins???

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Re: [algogeeks] seperate diff types of coins

2012-07-10 Thread payal gupta
@navin...Sorry didnt get you how come u were able to segregate all the
coins by the proposed method??


On 7/10/12, Navin Kumar algorithm.i...@gmail.com wrote:
 Minimum no. weighings:

 Divide 8 coins in group of 3, 3 and 2.

 For minimum weighsing group1 's total weight is x units(say) --FIrst
 weighing
 Groups 2nd total weights is y units  Second weighing.
 Lastly one more weighing among a unit and b unit coins.---3 rd weighing


 So minimum 3 weighing is required.



 On Tue, Jul 10, 2012 at 11:03 AM, payal gupta gpt.pa...@gmail.com wrote:

 You have 8 coins. 3 of them weigh x units, 3 y units, 1 a units and 1 b
 units. They are all mixed and look identical. What are the minimum no of
 weighings reqd to seperate the for types of coins???

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Re: [algogeeks] seperate diff types of coins

2012-07-10 Thread Navin Kumar
Let we have 8 coins named (for simplicity) x1, x2, x3, y1, y2, y3, a, b

According to your question 3 of them weigh x units : x1+x2+x3 =x

3 of them weigh y units: y1+y2+y3 =y

now consider the case when i grouped  (x1, x2 x3) in single group

(y1,y2 ,y3) in one group and (a,b) in one group.

Now we will measure weight of 1st group (x1, x2, x3) :It will give x then
we conclude that all elements in this group are those element whose total
weight is x :Type 1 element.

similarly for y.

and for the last group we will pick up any element and measure its weight
..its weight will be either a or b. Depending upon outcome we will
categorize them a and b.

So 3 weighing is required.

On Tue, Jul 10, 2012 at 7:05 PM, payal gupta gpt.pa...@gmail.com wrote:

 @navin...Sorry didnt get you how come u were able to segregate all the
 coins by the proposed method??


 On 7/10/12, Navin Kumar algorithm.i...@gmail.com wrote:
  Minimum no. weighings:
 
  Divide 8 coins in group of 3, 3 and 2.
 
  For minimum weighsing group1 's total weight is x units(say) --FIrst
  weighing
  Groups 2nd total weights is y units  Second weighing.
  Lastly one more weighing among a unit and b unit coins.---3 rd weighing
 
 
  So minimum 3 weighing is required.
 
 
 
  On Tue, Jul 10, 2012 at 11:03 AM, payal gupta gpt.pa...@gmail.com
 wrote:
 
  You have 8 coins. 3 of them weigh x units, 3 y units, 1 a units and 1 b
  units. They are all mixed and look identical. What are the minimum no of
  weighings reqd to seperate the for types of coins???
 
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 Groups
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[algogeeks] seperate diff types of coins

2012-07-09 Thread payal gupta


You have 8 coins. 3 of them weigh x units, 3 y units, 1 a units and 1 b 
units. They are all mixed and look identical. What are the minimum no of 
weighings reqd to seperate the for types of coins???

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